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CAT 2023 Slot 3 Solved Paper

Instructions [25 - 29]

An air conditioner (AC) company has four dealers - D1, D2, D3 and D4 in a city. It is evaluating sales performances of these dealers. The company sells two variants of ACs - Window and Split. Both these variants can be either Inverter type or Non-inverter type. It is known that of the total number of ACs sold in the city, 25% were of Window variant, while the rest were of Split variant. Among the Inverter ACs sold, 20% were of Window variant.
The following information is also known:
1. Every dealer sold at least two window ACs.
2. D1 sold 13 inverter ACs, while D3 sold 5 Non-inverter ACs.
3. A total of six Window Non-inverter ACs and 36 Split Inverter ACs were sold in the city.
4. The number of Split ACs sold by D1 was twice the number of Window ACs sold by it.
5. D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.
6. D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.
7. D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.

Question : 27

Total: 66

What was the total number of ACs sold by D2 and D4?

Your Answer:

Validate

Solution:

Let us assume, A is the total number of AC's sold
⇒ From the information that the total number of ACs sold in the city, 25% were of Window variant ⇒ Window AC′ s=A/4 and Split AC′ s=3A/4
Now, let us assume B is the total number of inverter A C's
⇒ From the information that among the Inverter ACs sold, 20% were of Window variant.⇒ Window Inverter AC′s=B/5 and Window Non-Inverter AC's =4B/5

Total (A)
Split (3A/4)		Window (A/4)
Inv (4B/5)	Non-Inv	Inv (B/5)	Non-Inv

From - Condition-3
⇒A/4−B/5=6 and 4B/5=36⇒B=46 and A=60

Total =60
Split = 45		Window = 15
Inv = 36	Non-Inv = 9	Inv = 9	Non-Inv = 6

Now, from condition-6
a) D1 & D4 sold "0" window Non-inverter ACs⇒ D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3⇒ D2 sold 4 and D3 sold 2 ACs of this type.
From condition-2
b) Let us assume, D1 sold "x" window inverter ACs ⇒ Number of split inverter ACs sold is 13−x
From condition-4
c) Number of split ACs sold by D1 will be " 2x "
From condition-5
d) Let us assume ' y ' is the number of window ACs sold by D3 & D4 ⇒ D2 sold 3y ACs of this type.
From condition-7
e) Let us assume ' z ' is the number of split inverter ACs sold by D3 and D4 ⇒ D2 sold 2z ACs of this type.
Let us use a,b,c,d, and e make a table:

D1 Total =
Split =		Window =x
Inv =13−x	Non-Inv =	Inv =x	Non-Inv =0

D2 Total =
Split =		Window =3y
Inv =2z	Non-Inv =	Inv =	Non-Inv =4

D3 Total =
Split =		Window =y
Inv =z	Non-Inv =3	Inv =	Non-Inv =2

D4 Total =
Split =		Window =y
Inv =z	Non-Inv =	Inv =	Non-Inv =0

We know that the total number of window ACs is 15
⇒ x+3y+y+y=15⇒x+5y=15, also x and y should be greater than or equal to 2 from condition -1
⇒ x=5 and y=2 is the only solution.
Filling this in the table:

D1 Total =
Split =		Window =5
Inv =8	Non-Inv =	Inv =5	Non-Inv =0

D2 Total =
Split =		Window =6
Inv =2z	Non-Inv =	Inv =2	Non-Inv =4

D3 Total =
Split =		Window =2
Inv =z	Non-Inv =3	Inv =0	Non-Inv =2

D4 Total =
Split =		Window =2
Inv =z	Non-Inv =	Inv =2	Non-Inv =0

Now, Number of split inverter ACs is 36
⇒8+2z+z+z=36⇒4z=28⇒=7.
Filling this and using (5), the number of split AC's sold by D1 is 2⋆5=10.

D1 Total =15
Split =10		Window =5
Inv =8	Non-Inv =2	Inv =5	Non-Inv =0

D2 Total =
Split =		Window =6
Inv =14	Non-Inv =	Inv =2	Non-Inv =4

D3 Total =12
Split =10		Window =2
Inv =7	Non-Inv =3	Inv =0	Non-Inv =2

D4 Total =
Split =		Window =2
Inv =7	Non-Inv =	Inv =2	Non-Inv =0

Total number of ACs sold by D2 and D4=60−D1−D3=60−15−12=33.

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